Numerical Analysis EE NCKU TienHao Chang Darby Chang
Numerical Analysis EE, NCKU Tien-Hao Chang (Darby Chang) 1
In the previous slide n Fixed point iteration scheme – what is a fixed point? – iteration function – convergence n Newton’s method – tangent line approximation – convergence n Secant method 2
In this slide n Accelerating convergence – linearly convergent – Newton’s method on a root of multiplicity >1 – (exercises) n Proceed to systems of equations – linear algebra review – pivoting strategies 3
2. 6 Accelerating Convergence 4
Accelerating convergence n Having spent so much time discussing convergence – is it possible to accelerate the convergence? n n How to speed up the convergence of a linearly convergent sequence? How to restore quadratic convergence to Newton’s method? – on a root of multiplicity > 1 5
Accelerating convergence Linearly convergence n Thus far, the only truly linearly convergent sequence – false position – fixed point iteration n Bisection method is not according to the definition 6
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Aitken’s Δ 2 -method n Substituting Eq. (2) into Eq. (1) n Substituting Eq. (4) into Eq. (3) n The above formulation should be a better approximation to p than pn 8
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Aitken’s Δ 2 -method Accelerated? which implies superanswer linearly convergence later 10
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Any Questions? About Aitken’s Δ 2 -method 12
Accelerating convergence Anything to further enhance? 13
Why not use p-head instead of p? 14
Steffensen’s method 15
Restoring quadratic convergence to Newton’s method 16
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Any Questions? 19
Two disadvantages n Both the first and the second derivatives of f are needed answer n Each iteration requires one more function evaluations 20
Any Questions? Chapter 2 Rootfinding (2. 7 is skipped) 21
Exercise 2010/4/21 9: 00 am Email to darby@ee. ncku. edu. tw or hand over in class. You may arbitrarily pick one problem among the first three, which means this exercise contains only five problems. 22
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(Programming) 27
Chapter 3 Systems of Equations 28
Systems of Equations Definition 29
3. 0 Linear Algebra Review (vectors and matrices) 30
Matrix Definitions 31
Any Questions? m, n, m, i, j, EQUAL, SUM, SCALAR MULTIPLICATION, PRODUCT… 32
The Inverse Matrix (cannot be skipped) 33
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Any questions? question answer 35
The Determinant (cannot be skipped, too) 36
cofactor 37
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Link the concepts – All these theorems will be extremely important throughout this chapter n Nonsingular matrices n Determinants n Solutions of linear systems of equations 39
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(Hard to prove) 41
Any Questions? 3. 0 Linear Algebra Review 42
3. 1 Gaussian Elimination (I suppose you have already known it) 43
An application problem 44
n I 1 -I 2 -I 3=0 n I 2 -I 4 -I 5=0 n I 3+I 4 -I 6=0 n 2 I 3+I 6=7 n I 2+2 I 5=13 n -I 2+2 I 3 -3 I 4=0 45
Following Gaussian elimination 46
Any Questions? Gaussian elimination 47
Gaussian elimination Operation Counts 48
Operation Counts Comparison n Gaussian elimination – forward elimination – back substitution n Gauss-Jordan elimination n Compute the inverse matrix 49
3. 2 Pivoting Strategy 50
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Compare to x 1=1, x 2=7, x 3=1 53
Pivoting strategy n n n To avoid small pivot elements A scheme for interchanging the rows (interchanging the pivot element) Partial pivoting 54
http: //thomashawk. com/hello/209/1017/1024/Jackson%20 Running. jpg In action 55
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Compare to x 1=1, x 2=7, x 3=1 57
Any Questions? 58
From the algorithm view n How to implement the interchanging operation? – changehint implicitly n Introduce a row vector r – each time a row interchange is required, answer we need only swap the corresponding elements of the vector – number of operations from 3 n to 3 59
http: //thomashawk. com/hello/209/1017/1024/Jackson%20 Running. jpg In action 60
Without pivoting 61
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n x = [1. 000, -0. 9985, 0. 9990, -1. 000]T – exact solution x = [1, -1, 1, -1]T – no r x = [1. 131, -0. 7928, 0. 8500, -0. 9987]T 64
Scaled Partial Pivoting 65
Scaled partial pivoting An example 66
Any Questions? 67
Scaled partial pivoting A blind spot of partial pivoting answer 68
Scaled partial pivoting 69
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http: //thomashawk. com/hello/209/1017/1024/Jackson%20 Running. jpg In action 71
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n x = [1. 000, -1. 000]T – exact solution x = [1, -1, 1, -1]T – no s x = [1. 000, -0. 9985, 0. 9990, -1. 000]T – no r x = [1. 131, -0. 7928, 0. 8500, -0. 9987]T 74
Any Questions? 3. 2 Pivoting Strategy 75
- Slides: 75